標題:
一條數學問題~~一條數學題
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發問:
y+1=y/tan30' 點做~~？ 更新: = =各位，你地的解釋好難體得明，可以寫好D嗎?
最佳解答:
y+1=y/tan30' Sol tan30度=1/sqrt(3) y+1=y/(1/sqrt(3)) y+1=sqrt(3)y sqrt(3)y－y=1 y(sqrt(3)－1)=1 y=1/[sqrt(3)－1] y=[sqrt(3)+1/[(sqrt(3)+1))sqrt(3)－1)] y=(sqrt(3)+1)/2
其他解答:
y+1=y/tan30 since tan 30 = 1/√3 so the equation can be rewrite as the following : y+1=y/(1/√3) =>y+1=√3 y =>(1-√3)y=-1 =>y=1/(√3-1)=(√3+1)/(3-1)=(√3+1)/2//|||||.........y + 1 = y/tan 30 .........y + 1 = y x 3^0.5 ......1 + 1/y = 3^0.5 ............1/y = (3^0.5) - 1 ...............y = 1.366025~~|||||y+1=y/tan30' tan30=1/sqrt(3) y+1=y/tan30 => y+1=y/(1/sqrt(3)) => y+1=sqrt(3)y =>1=[sqrt(3)-1]y =>y=1/[sqrt(3)-1] =>y=[sqrt(3)+1]/[sqrt(3)-1][sqrt(3)+1] =>y=[sqrt(3)+1]/(3-1) =>y=[sqrt(3)+1]/2 or 1.366 2009-06-20 21:55:15 補充： y+1=y/tan30 y+1=y/0.5774 (你用計算機就知道tan30=0.5774) y+1=1.732y y+1-y=1.732y-y 1=0.732y 1/0.732=0.732y/0.732 1.366=y y=1.366
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